Unformatted text preview: . .tlKll)(286 K) V == =l1L I PI (0.747 atm) What is V 2 ? n2RT2 (0.35 .tlKll)(0.08206 L . atmJK . .tlKll)(329 K) V 2 =p;= (1.18 atrn) = 8.0 L What is the change in volume L1 V? • Ll V = V 2VI = 8.0 L 11 L =3 L The change in volume is negative because the volume decreases. Note: For this problem (unlike Example 5.9), the pressures must be converted from torrs to atmospheres as required by the atmospheres part of the units for R since each volume was found separately, and the conversion factor does not cancel. SEE EXERCISE 5.61 5.4 ~ Gas Stoichiometry Suppose we have I mole of an ideal gas at O°C (273.2 K) and 1 atm. From the ideal gas law, the volume of the gas is given by nRT (1.000 .tlKll)(0.08206 L . atmJK . .tlKll)(273.2 K) V = = = 22.42 L P 1.000 atrn...
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 Spring '10
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 Organic chemistry, 3 L, 11 L

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